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A 1.0 kg object is attached to a string 0.50 m. It is twirled in a horizontal circle above the ground at a speed of 5.0 m/s. A ball is moving on a circular path. The ball is currently at the top right position of the circle. Point a is above and to the left of the ball. Point b is above and to the right of the ball. Point c is down and to the right of the ball. Point d is down and to the left of the ball and inside the circle near the center. What is the magnitude of the centripetal force acting on the object? 2.5 N 10. N 25 N 50. N

Respuesta :

Kalahira
50 N The centripetal force upon an object is expressed as F = mv^2/r So let's substitute the known values and calculate F = mv^2/r F = 1.0 kg * (5.0 m/s)^2 / 0.5 m F = 1.0 kg * 25 m^2/s^2 / 0.5 m F = 25 kg*m^2/s^2 / 0.5 m F = 50 kg*m/s^2 F = 50 N So the answer is 50 N which matches one of the available choices.
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